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Vector Algebra

Question
CBSEENMA12032947

Form the differential equation of the family of curves
straight y equals Ax plus straight B over straight x
where A and B are constants.

Solution

The given equation is
                      straight y equals Ax plus straight B over straight x                                     ...(1)
therefore space space space dy over dx space equals space straight A minus straight B over straight x squared                                              ...(2)
rightwards double arrow space space fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space fraction numerator 2 straight B over denominator straight x cubed end fraction                                                  ...(3)
From (3), B = straight x cubed over 2 fraction numerator straight d squared straight y over denominator dx squared end fraction
From  (2),  dy over dx space equals space straight A minus straight x over 2 fraction numerator straight d squared straight y over denominator dx squared end fraction space space rightwards double arrow space space space straight A space equals space straight x over 2 fraction numerator straight d squared straight y over denominator dx squared end fraction plus dy over dx
Putting value of A and B in (1), we get,
                                straight y space equals straight x squared over 2 fraction numerator straight d squared straight y over denominator dx squared end fraction plus straight x dy over dx plus straight x squared over 2 fraction numerator straight d squared straight y over denominator dx squared end fraction space space space or space space space straight x squared fraction numerator straight d squared straight y over denominator dx squared end fraction plus straight x dy over dx minus straight y space equals 0
which is the required differential equation.